1) s="\\sigma\/\\sqrt n"
n=10,so
s=3.5/3.16=1.11
n=47,so
s=3.5/6.855=0.51
As the sample size is increased, standard deviation of the sample mean decreases.
2) n=35,a=1,b=3
"\\mu"X=(b+a)/2=2
"\\sigma"X= "\\sqrt{((b-a+1)^2-1)\/12}" =0.8165
P(2.1<X<2.4)=P((2.1-2)/(0.8165/"\\sqrt{ 35}") )<Z<((2.4-2)/(0.8165/"\\sqrt {35}" ))=P(0.72<Z<2.9)=P(z=2.9)-P(z=0.72)=0.9981-0.7642=0.2339
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